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The Congruent Number Problem

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The Congruent Number Problem THE CONGRUENT NUMBER PROBLEM KEITH CONRAD 1. Introduction A right triangle is called rational when its legs and hypotenuse are all rational numbers. Examples of rational right triangles include Pythagorean triples like (3, 4, 5). We can scale such triples to get other rational right triangles, like (3/2, 2, 5/2). Of course, usually√ when two sides are rational the third side is not rational, such as the (1, 1, 2) right triangle. Any rational right triangle has a rational area, but not all (positive) ratio- nal numbers can occur as the area of a rational right triangle. For instance, no rational right triangle has area 1. This was proved by Fermat. The ques- tion we will examine here is: which rational numbers occur as the area of a rational right triangle? Definition 1.1. A positive rational number n is called a congruent number if there is a rational right triangle with area n: there are rational a, b, c > 0 such that a2 + b2 = c2 and (1/2)ab = n. This archaic use of the word congruent has nothing to do (directly) with congruences in modular arithmetic. The etymology will be explained in Section 3. The history of congruent numbers can be found in [2, Chap. XVI], where it is indicated that an Arab manuscript called the search for congruent numbers the “principal object of the theory of rational right triangles.” Example 1.2. The number 6 is congruent, being the area of the (3, 4, 5) right triangle. Example 1.3. The (9, 40, 41) right triangle has area 180 = 5·62, so dividing the lengths by 6 produces the rational right triangle (3/2, 20/3, 41/6) with area 5. That is, 5 is a congruent number. Example 1.4. The right triangle with sides (175, 288, 337) has area 25200 = 7·602, so scaling by 60 produces the right triangle (35/12, 24/5, 337/60) with area 7. Thus 7 is a congruent number. The congruent number problem asks for a description of all congruent numbers. Since scaling a triangle changes its area by a square factor, and every rational number can be multiplied by a suitable rational square to become a squarefree integer (e.g., 18/7 = 32 · 2/7, so multiplying by (7/3)2 produces the squarefree integer 14), we can focus our attention in the con- gruent number problem on squarefree positive integers. For instance, to 1 2 KEITH CONRAD say 1 is not a congruent number means no rational square is a congruent number. When n is squarefree in Z+, to show n is a congruent number we just need to find an integral right triangle whose area has squarefree part n. Then writing the area as m2n shows scaling the sides by m produces a rational right triangle with area n. 2. A bad algorithm There is a parametric formula for primitive Pythagorean triples, and using it we will make a small list of squarefree congruent numbers. Any primitive triple (with even second leg) is (k2 − `2, 2k`, k2 + `2) where k > ` > 0, (k, `) = 1, and k 6≡ ` mod 2. In Table 1 we list such primitive triples where k + ` ≤ 9. The squarefree part of the area is listed in the last column. Each number in the fourth column is a congruent number and each number in the fifth column is also a congruent number. The final row of the table explains how Example 1.3 is found. k ` (a, b, c) (1/2)ab Squarefree part 2 1 (3, 4, 5) 6 6 4 1 (15, 8, 17) 60 15 3 2 (5, 12, 13) 30 30 6 1 (35, 12, 37) 210 210 5 2 (21, 20, 29) 210 210 4 3 (7, 24, 25) 84 21 8 1 (63, 16, 65) 504 126 7 2 (42, 28, 53) 630 70 5 4 (9, 40, 41) 180 5 Table 1. Congruent Numbers Notice 210 shows up twice in Table 1. Do other numbers which occur once also occur again? We will return to this question later. Table 1 can be extended according to increasing values of k + `, and any squarefree congruent number eventually shows up in the last column, e.g., the triangle (175, 288, 337) from Example 1.4, whose area has squarefree part 7, occurs at k = 16 and ` = 9. Alas, the table is not systematic in the appearance of the last column: we can’t tell by building the table when any particular number should occur, if at all, in the last column, so this method of generating (squarefree) congruent numbers is not a good algorithm. For instance, 53 is a congruent number, but it shows up for the first time when k = 1873180325 and ` = 1158313156. (The corresponding right triangle has area 53 · 2978556542849787902.) Tabulations of congruent numbers can be found in Arab manuscripts from the 10th century, and 5 and 6 appear there. Fibonacci discovered in the 13th century that 7 is congruent and he stated that 1 is not congruent (that is, no THE CONGRUENT NUMBER PROBLEM 3 rational right triangle has area equal to a perfect square). The first accepted proof is due to Fermat in the mid-1600s. Theorem 2.1 (Fermat). The numbers 1 and 2 are not congruent. Proof. In Table 2, the first two columns show how to convert the sides (a, b, c) of a rational right triangle with area 1 into a positive rational solution of the equation y2 = x4 − 1 and conversely. (These correspondences are not inverses, but at least they show a positive rational solution in the first column leads to a positive rational solution in the second column, and conversely.) The last two columns give a (bijective) correspondence between rational right triangles with area 2 and positive rational solutions of y2 = x4 + 1. So showing 1 and 2 are not congruent numbers is the same as showing the equations y2 = x4 ± 1 don’t have solutions in positive rational numbers. a2 + b2 = c2, y2 = x4 − 1 a2 + b2 = c2, y2 = x4 + 1 1 1 2 ab = 1 2 ab = 2 x = c/2 a = y/x x = a/2 a = 2x y = |a2 − b2|/4 b = 2x/y y = ac/4 b = 2/x c = (x4 + 1)/xy c = 2y/x Table 2 A positive rational solution to y2 = x4 ± 1 can be turned into a positive integral solution of w2 = u4 ± v4 by clearing a common denominator. We can go in reverse by dividing by v4. That 1 and 2 are not congruent is therefore the same as the equations w2 = u4 ± v4 having no positive integer solutions. The reader is referred to [1, pp. 252–256] for a proof by descent that w2 = u4 ± v4 has no positive integer solutions. (Fermat discovered his method of descent on exactly this problem.) Remark 2.2. Fermat considered the equations w2 = u4 ± v4 merely as auxiliary tools to get the proof of Theorem 2.1, but nowadays these equations are more prominent than Theorem 2.1 itself since the equations w2 = u4 ±v4 are related to the exponent 4 case of Fermat’s Last Theorem. √ √ Theorem 2.1√ leads√ to two weird proofs that 2 is irrational. If 2 were rational then 2, 2, and 2 are the sides of a rational right triangle with area√ 1. This is a contradiction of 1 not√ being a congruent number! Similarly, if 2 were rational then 2, 2, and 2 2 would be the sides of a rational right triangle with area 2, but 2 is not a congruent number! Fermat showed 3 is √not a congruent number, and the reader is invited to deduce from this that 3 is irrational. 4 KEITH CONRAD 3. Relation to Arithmetic Progressions of Three Squares The three squares 1, 25, 49 form an arithmetic progression with common difference 24. The squarefree part of 24 is 6. This is related to 6 being a congruent number, by the following theorem. Theorem 3.1. Let n > 0. There is a one-to-one correspondence between right triangles with area n and 3-term arithmetic progressions of squares with common difference n: the sets {(a, b, c) : 0 < a < b < c, a2 + b2 = c2, (1/2)ab = n} and {(r, s, t) : 0 < r < s < t, s2 − r2 = n, t2 − s2 = n} are in one-to-one correspondence by (a, b, c) 7→ ((b − a)/2, c/2, (b + a)/2), (r, s, t) 7→ (t − r, t + r, 2s). Proof. It is left to the reader to check the indicated functions take values in the indicated sets, and that the correspondences are inverses of one another: if you start with an (a, b, c) and make an (r, s, t) from it, and then form an (a0, b0, c0) from this (r, s, t), you get back the original (a, b, c). Similarly, starting with an (r, s, t), producing an (a, b, c) from it and then producing 0 0 0 an (r , s , t ) from that returns the same (r, s, t) you started with. When n > 0 is rational, the correspondence in Theorem 3.1 preserves rationality: (a, b, c) is a rational triple if and only if (r, s, t) is a rational triple. Therefore n is congruent if and only if there is a rational square s2 such that s2 − n and s2 + n are also squares. Note the correspondence in Theorem 3.1 involves not the squares in arithmetic progression but their positive rational square roots r, s, and t. Example 3.2.
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